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  • For a reaction:  <img alt="\frac{5}{2}\mathrm A\rightarrow \mathrm{products}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B5%7D%7B2%7D%5Cmathrm%20A%5Crightarrow%20%5Cma

For a reaction:
 \frac{5}{2}\mathrm A\rightarrow \mathrm{products}   
\mathrm{ r=k[A]^{0}}  
initial concentration of A = A0. Find the concentration of A after time t.

Option: 1

\mathrm{A = A_{0}-\frac{5}{2}kt}


Option: 2

\mathrm{A = A_{0}-\frac{2}{5}kt}


Option: 3

\mathrm{A = A_{0}-kt}


Option: 4

\mathrm{A = A_{0}}


Answers (1)

best_answer

Given reaction:
 \frac{5}{2}\mathrm A\rightarrow \mathrm{products}    

It will be a zero-order reaction due to rate is not depending on reactant concentration.

For zero-order reaction, the rate equation is given as follows:

\mathrm{A\: =\: A_{o}-kt}

According to the question, we have given:

\mathrm{r\: =\: k[A]^{0}\: =\: -\frac{1}{\frac{5}{2}}\frac{dA}{dt}}

\mathrm{-\frac{1}{\frac{5}{2}}\frac{dA}{dt}\: =\: k\quad\quad\quad\quad\quad(Since\: [A]^{0}\: =\: 1)}

Now, integrating both sides we get:

\mathrm{[A]^{A}_{A_{o}}\: =\: -\frac{5}{2}k(t)^{t}_{o}}

\mathrm{A-A_{o}\: =\: -\frac{5}{2}kt}

\mathrm{\mathbf{A\: =\: A_{o}-\frac{5}{2}kt}}
This is the concentration of A after time 't'.

Therefore, option(1) is correct
 

Posted by

Suraj Bhandari

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